Problem: Let $g(x)=\dfrac{1}{x^4}-\dfrac{3}{x^2}+x^3$. $g'(x)=$
The strategy We can first rewrite each rational term of $g$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting rational terms as negative powers $\begin{aligned} g(x)&=\dfrac{1}{x^4}-\dfrac{3}{x^2}+x^3 \\\\ &=x^{-4}-3x^{-2}+x^3 \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}(x^{-4}-3x^{-2}+x^3) \\\\ &=\dfrac{d}{dx}(x^{-4})-3\dfrac{d}{dx}(x^{-2})+\dfrac{d}{dx}(x^3) \\\\ &=-4x^{-5}-3(-2)x^{-3}+3x^2 \\\\ &=-\dfrac{4}{x^5}+\dfrac{6}{x^3}+3x^2 \end{aligned}$ In conclusion, $g'(x)=-\dfrac{4}{x^5}+\dfrac{6}{x^3}+3x^2$.